Skip to main content

Section 3.1 Banach spaces

Definition 3.1.1. Banach space.
A Banach space is a normed vector space which is complete with respect to the induced metric space structure.
As mentioned in Theorem 2.5.3, in finite dimension there is a single (modulo isomorphisms) normed vector space over the real (or complex) numbers and that normed vector space is Banach.

In infinite dimension, the situation is quite different. Below we illustrate, in some detail, an important example, namely the set \(C^\infty([0,1])\) of smooth functions on the unit interval. In particular, we exhibit two families of countably many norms on \(C^\infty([0,1])\) all pairwise inequivalent. Moreover, \(C^\infty([0,1])\) is not Banach with respect to any of them.

Subsection 3.1.1 Example: norms on \(C^\infty([0,1])\)

The elements of \(C^\infty([0,1])\) are the functions \(f:[0,1]\to\bR\) that are differentiable infinitely many times. In particular, this implies that those functions and each of their derivatives are continuous. The fact that \(C^\infty([0,1])\) is a linear space is an immediate consequence of the linearity of the first derivative operator. The fact that \(C^\infty([0,1])\) is infinite-dimensional is due to the fact that each monomial \(x^n\) is linearly independent on all other monomials, namely \(C^\infty([0,1])\) contains infinitely many elements which are linearly independent on each other.

The \(C^0\) norm. Set
\begin{equation*} \|f\|_{C^0} = \sup_{x\in[0,1]}|f(x)|. \end{equation*}
One can easily verify that \(\|\cdot\|_{C^0}\) is indeed a norm on \(C^\infty([0,1])\) (see Exercise 3.8.1).

Is \(C^\infty([0,1])\) a Banach space with respect to \(\|\cdot\|_{C^0}\text{?}\) By definition, a sequence \(f_n\) converges to \(f\) in the \(\|\cdot\|_{C^0}\) norm if and only if \(f_n\) converges uniformly to \(f\text{.}\) Then, by the uniform limit theorem, we known that, at the very least, \(f\in C^0([0,1])\text{.}\)

Can we prove that actually \(f\in C^\infty([0,1])\text{?}\) The picture generated by the code below is enough to understand that this is not the case.
More analytically, set
\begin{equation} f_n(x) = \sqrt{(x-0.5)^2+\frac{1}{n}}.\label{eq-sqrt}\tag{3.1.1} \end{equation}
Then, since
\begin{equation*} 0\leq\sqrt{(x-0.5)^2+\frac{1}{n}}-(x-0.5) = \frac{1/n}{\sqrt{(x-0.5)^2+\frac{1}{n}}+(x-0.5)}\leq\frac{1}{n}, \end{equation*}
\(f_n\) converges, in the \(C^0\) norm, to \(g(x)=|x-0.5|\text{,}\) which is not even \(C^1\text{.}\)

Hence, \(C^\infty([0,1])\) is not complete under the \(C^0\) norm. What is the completion of this metric space? The Weierstrass theorem grants that every \(C^0\) function in an interval can be uniformly approximated by polynomials, which are \(C^\infty\) functions, so the completion of \(C^\infty([0,1])\) with respect to \(\|\cdot\|_{C^0}\) is \(C^0([0,1])\text{.}\)

In particular, this shows that \((C^0([0,1]),\|\cdot\|_{C^0})\) is a Banach space.

The \(C^1\) norm. Set now
\begin{equation*} \|f\|_{C^1} = \|f\|_{C^0} + \|f'\|_{C^0}. \end{equation*}
One can easily verify, similarly to the case above, that \(\|\cdot\|_{C^1}\) is a norm on \(C^\infty([0,1])\) (see Exercise 3.8.1).

Is \(C^\infty([0,1])\) a Banach space with respect to \(\|\cdot\|_{C^1}\text{?}\) A sequence \(f_n\) converges to \(f\) in the \(\|\cdot\|_{C^1}\) norm if and only if \(f_n\) converges uniformly to \(f\) and \(f'_n\) converges uniformly to \(f'\text{.}\) Consider now a Cauchy sequence \(f_n\) with respect to the \(C^1\) norm. Then both \(f_n\) and \(f'_n\) are Cauchy with respect to the \(C^0\) norm. Hence, since \((C^0([0,1]),\|\cdot\|_{C^0})\) is Banach, there are \(f,g\in C^0([0,1])\) with \(f_n\to f\) and \(f'_n\to g\) in \(C^0\text{.}\) Since each \(f_n\) is \(C^1\) then, by the fundamental theorem of calculus,
\begin{equation*} f_n(x) - f_n(0) = \int_0^x f'_n(s)ds. \end{equation*}
\begin{equation*} f_n(x) - f_n(0) = \int_0^x f'_n(s)ds. \end{equation*}
Since the \(f'_n\) converge uniformly to \(g\text{,}\) then, by taking the limit, we have that
\begin{equation*} f(x) - f(0) = \int_0^x g(s)ds, \end{equation*}
so that \(g=f'\text{.}\) Hence \(f\) is \(C^1\) and \(f_n\to f\) with respect to \(\|\cdot\|_{C^1}\text{.}\) Again by Wierstrass theorem, we have that every \(C^1\) function can be approximated by smooth functions, which means that the completion of \(C^\infty([0,1])\) with respect to \(\|\cdot\|_{C^1}\) is \(C^1([0,1])\text{.}\)

In particular, this shows that \((C^1([0,1]),\|\cdot\|_{C^1})\) is a Banach space.

The \(C^k\) norm. Set now, for an integer \(k>1\text{,}\)
\begin{equation*} \|f\|_{C^k} = \|f\|_{C^0} + \|f'\|_{C^0} + \dots + \|f^{(k)}\|_{C^0}. \end{equation*}
A discussion analogous to the one we did above will show that \(C^\infty([0,1])\) is not complete with respect to \(\|\cdot\|_{C^1}\) and its completion gives precisely the Banach space \((C^k([0,1]),\|\cdot\|_{C^k})\text{.}\)

Each of these \(C^k\) norms, \(k\geq0\text{,}\) are therefore inequivalent on \(C^\infty([0,1])\text{,}\) since the completion of \(C^\infty([0,1])\) with respect to each of them gives different Banach spaces.

The \(L^1\) norm. Now we turn to norms defined via integrals rather than derivatives. Given a smooth function \(f\text{,}\) we set
\begin{equation*} \|f\|_{L^1} = \int_0^1|f(s)|\;ds, \end{equation*}
where the integral is a Riemann integral. This function is well-defined on \(C^\infty([0,1])\) because \([0,1]\) has finite measure and every continuous function over a compact is bounded. Moreover, Minkowski's inequality (see ???) grants that \(\|\cdot\|_{L^1}\) is a norm on \(C^\infty([0,1])\text{.}\)

Is \(C^\infty([0,1])\) a Banach space with respect to \(\|\cdot\|_{L^1}\text{?}\) Once again, it is easy to see that this is not the case. For instance, as shown in ???, in this norm a sequence of smooth functions can converge to a step function, that is not even \(C^0\text{.}\)

We denote by \(L^1([0,1])\) the completion of \(C^\infty([0,1])\) with respect to \(\|\cdot\|_{L^1}\text{.}\) This is the space of all Lebesgue-integrable functions on \([0,1]\). In other words, this can be taken as an alternate definition of the Lebesgue integral.

The reason why we started with the Riemann integral and ended up, in the completion, with the Lebesgue integral is precisely that the Riemann integral does not behave nicely under limits. For instance, let \(a_n\) be a sequence containing all rational numbers in \([0,1]\) and no other number. Then, each function \(f_n\) which is one at every point \(a_k\) with \(k\leq n\) and zero otherwise is Riemann-integrable but, in the limit for \(n\to\infty\text{,}\) \(f_n\) converges pointwise to the Dirichlet function \(1_\bQ\text{,}\) that is not Riemann-integrable. On the contrary, \(1_\bQ\) is Lebesgue-integrable and \(\lim\int_{[0,1]}f_n(x) dx=\int_{[0,1]}(\lim f_n(x)) dx\text{.}\)

The \(L^p\) norm. For a given real number \(p\geq1\) and an integrable function \(f\) on \([0,1]\text{,}\) we set
\begin{equation*} \|f\|_{L^p} = \left[\int_0^1|f(s)|^p\;ds\right]^{1/p}. \end{equation*}
We discussed above the case \(p=1\text{.}\) Even for \(p>1\text{,}\) Minkowski's inequality grants that \(\|\cdot\|_{L^p}\) is a norm -- note that, on the contrary, \(\|\cdot\|_{L^p}\) is not a norm for \(p\in(0,1)\text{.}\)

We denote by \(L^p([0,1])\) the completion of \(C^\infty([0,1])\) with respect to \(\|\cdot\|_{L^p}\text{.}\) Which functions live in \(L^p([0,1])\text{?}\)

The following argument shows that \(L^p([0,1])\) contains every Lebesgue-integrable function whose \(L^p\) norm is finite. Let indeed \(f\in L^1([0,1])\) be such that \(\|f\|_{L^p}<\infty\) and, for \(M>0\text{,}\) define its \(M\)-truncation as \(f_M = f\cdot 1_{|f|\leq M}\text{,}\) where \(1_{|f|\leq M}\) is equal to 1 for \({|f|\leq M}\) and to 0 otherwise. Since \([0,1]\) has finite measure and \(\|f\|_{L^p}<\infty\text{,}\)
\begin{equation*} \|f - f_M\|_{L^p}^p = \int_{|f|>M} |f|^p \to 0 \quad \text{as } M\to\infty. \end{equation*}
Now, choose \(M\) so that \(\|f-f_M\|_{L^p} < \varepsilon/2\) and note that \(f_M \in L^1\text{.}\) Since \(L^1([0,1])\) is the completion of \(C^\infty\text{,}\) for every \(\delta>0\text{,}\) there exists \(\psi \in C^\infty([0,1])\) such that \(\|f_M - \psi\|_{L^1} < \delta\text{.}\) Set \(A = \max\{M, \|\psi\|_{C^0}\}\text{.}\) Notice that, if \(h\in L^1([0,1])\) is such that \(|h|\le A\text{,}\) then
\begin{equation*} |h|^p = |h|^{p-1} \cdot |h| \le A^{p-1} |h|, \end{equation*}
so that
\begin{equation*} \|h\|_{L^p}^p \le A^{p-1} \|h\|_{L^1}. \end{equation*}
Applying this to \(h=f_M-\psi\text{,}\) we obtain that
\begin{equation*} \|f_M - \psi\|_{L^p}^p \le A^{p-1} \|f_M - \psi\|_{L^1}\le A^{p-1}\delta. \end{equation*}
Now, choose \(\delta\) so small that \(A^{p-1}\delta < (\varepsilon/2)^p\text{.}\) Then \(\|f_M - \psi\|_{L^p} < \varepsilon/2\text{,}\) so that
\begin{equation*} \|f - \psi\|_{L^p} \le \|f - f_M\|_{L^p} + \|f_M - \psi\|_{L^p} < \varepsilon/2 + \varepsilon/2 = \varepsilon. \end{equation*}

In particular, this means that one can equivalently define \(L^p([0,1])\) as the set
\begin{equation*} L^p([0,1]) = \{f: f\text{ is Lebesgue-integrable on }[0,1]\text{ and }\|f\|_{L^p}<\infty\}. \end{equation*}

Each of these \(L^p\) norms, \(p\geq1\text{,}\) are inequivalent on \(C^\infty([0,1])\) since the corresponding completions \(L^p([0,1])\) are non-isomorphic Banach spaces (the fact that \(L^p\) is not isomorphic, as a Banach space, to \(L^q\) for \(p\neq q\) is non-trivial and we will not prove it in this book; the interested reader can look at the type and cotype of an \(L^p\) space).

So, is \(C^\infty([0,1])\) a Banach space with respect to some norm? This question should be specified more precisely. In principle yes, it can be done, but such a Banach space structure won't be compatible with the topology one would expect on \(C^\infty([0,1])\text{.}\) For instance, one feature we want from a topology on \(C^\infty([0,1])\) is that the derivative operator be continuous.

Let us consider first the operator \(\frac{d}{dx}:C^1([0,1])\to C^0([0,1])\text{.}\) Note that, by linearity, it is sufficient to check the continuity of a linear operator on sequences converging to zero (see Section ??? for more details). So, assume that \(f_n\to0\) in \(C^1([0,1])\text{.}\) We need to show that \(f'_n\to0\) in \(C^0([0,1])\text{.}\) This is obvious since \(f_n\to0\) in \(C^1([0,1])\) means that \(\|f_n\|_{C^0}+\|f'_n\|_{C^0}\to0\text{.}\) Hence, \(\frac{d}{dx}\) is continuous on \(C^1([0,1])\text{.}\)

We want the same to be true for the operator \(\frac{d}{dx}:C^\infty([0,1])\to C^\infty([0,1])\text{.}\) In this case, though, the source and target spaces are one and the same and so we would need to find a norm \(\|\cdot\|\) with the property that \(\|f_n\|\to0\) implies \(\|f'_n\|\to0\text{.}\) This, though, is impossible for the following reason. Set \(e_n(x)=e^{nx}\text{,}\) so that \(e'_n(x)=ne_n(x)\text{,}\) and set \(\|e_n\|=c_n\text{.}\) Then, although the sequence \(f_n=\frac{1}{\sqrt{n}}\frac{e_n}{c_n}\) converges to zero, the sequence of its derivatives \(f'_n=\sqrt{n}\frac{e_n}{c_n}\) diverges.

Hence, any Banach space structure on \(C^\infty([0,1])\) will make the first derivative discontinuous. On the other side, the \(C^k\) norms can be used to write a well-defined distance function on \(C^\infty([0,1])\) as follows:
\begin{equation*} d(f,g) = \sum_{k=0}^\infty \frac{1}{2^k}\frac{\|f-g\|_{C^k}}{1+\|f-g\|_{C^k}}. \end{equation*}
An equivalent, but more widely used, distance function is given by
\begin{equation*} d(f,g) = \sum_{k=0}^\infty \frac{1}{2^k}\frac{p_k(f-g)}{1+p_k(f-g)}, \end{equation*}
where
\begin{equation*} p_k(f) = \|f^{(k)}\|_{C^0}. \end{equation*}
Notice that each \(p_k\text{,}\) \(k>0\text{,}\) is only a semi-norm (as opposed to a norm) on \(C^\infty([0,1])\text{,}\) namely for every \(k>0\) there are non-zero elements \(f\in C^\infty([0,1])\) with \(p_k(f)=0\text{.}\)

Let us verify that these are distances. Since \(d(f,g)=0\) implies \(\|f-g\|_{C^0}=0\text{,}\) then it also implies \(f=g\text{.}\) The symmetry is also obvious. The triangular inequality is a consequence of the fact that the function
\begin{equation*} \varphi(t)=\frac{t}{1+t} \end{equation*}
is monotonically increasing in \(t\) and subadditive, namely \(\varphi(a+b)\leq\varphi(a)+\varphi(b)\text{,}\) for all \(a,b\geq0\text{.}\) Hence, for each \(k\geq0\) and \(h\in C^\infty([0,1])\text{,}\)
\begin{equation*} \varphi(p_k(f-g)) \leq \varphi(p_k(f-h)+p_k(h-g))\leq \varphi(p_k(f-h)) + \varphi(p_k(h-g)). \end{equation*}

One can verify the equivalence of the two distances above by checking that they give rise to the same converging sequences. If \(f_n\to g\) w.r.t. the first distance function, then necessarily \(\|f_n-g\|_{C^k}\to0\) for all \(k>0\text{.}\) In turn, this means that \(\|f^{(k)}_n-g^{(k)}\|_{C^0}\to0\) for all \(k>0\text{,}\) and viceversa.

Assume now that \(f_n\) is Cauchy in the metric space \((C^\infty([0,1]),d)\text{.}\) Then \(f_n\) is a Cauchy sequence also in each space \(C^k([0,1])\) and so if converges to a function that is \(C^k\) for every \(k\text{,}\) namely it is \(C^\infty\text{.}\) Hence, \((C^\infty([0,1]),d)\) is complete.

Notice that, with respect to this topology, the first derivative operator is continuous. Indeed, if \(f_n\to0\) in \((C^\infty([0,1]),d)\text{,}\) it means that
\begin{equation*} d(f_n,0) = \sum_{k=0}^\infty \frac{1}{2^k}\frac{p_k(f_n)}{1+p_k(f)}\to0, \end{equation*}
which is possible if and only if \(\|f_n\|_{C^k}\to0\) for every \(k\geq0\text{.}\) In turn, this is possible if and only if \(\|f'_n\|_{C^k}\to0\) for every \(k\geq0\text{,}\) which means that \(f'_n\to0\) in \((C^\infty([0,1]),d)\text{.}\)

A space whose topology is given by a countable family of semi-norms \(p_k\text{,}\) such as \(C^\infty([0,1])\text{,}\) and is Hausdorff and a complete metric space with the metric above (notice that such a metric is always translation-invariant) is called a Fréchet space.

Although it won't be relevant in this book, for completeness sake we mention that, for \(p\in(0,1)\text{,}\) the function \(\|\cdot\|_{L^p}\) is not a norm but the function
\begin{equation*} d_{L^p}(f,g) = \int_{[0,1]}|f(s)-g(s)|^p\;ds \end{equation*}
is a distance function. Indeed, the only non-trivial axiom to verify is the triangular inequality:
\begin{equation*} \int_{[0,1]}|f(s)-g(s)|^p\;ds\leq\int_{[0,1]}|f(s)-h(s)|^p\;ds + \int_{[0,1]}|h(s)-g(s)|^p\;ds\. \end{equation*}
We set \(a=f-h\) and \(b=h-g\text{,}\) so that the relation becomes
\begin{equation*} \int_{[0,1]}|a(s)+b(s)|^p\;ds\leq\int_{[0,1]}|a(s)|^p\;ds + \int_{[0,1]}|b(s)|^p\;ds. \end{equation*}
This holds because actually \(|a(s)+b(s)|^p\leq |a(s)|^p+|b(s)|^p\) pointwise for every \(p\in(0,1]\text{.}\) Indeed, let \(t=a/b\text{.}\) Then, by factoring out $a$ and noticing that the relation is trivially true when \(a(s)=0\text{,}\) we reduce to
\begin{equation*} (1+t)^p\leq 1+t^p. \end{equation*}
A direct study of the function \(1+t^p-(1+t)^p\) shows that it is non-negative for \(t\geq0\) and \(p\in(0,1]\text{,}\) which proves the claim.

The metric space \((C^\infty([0,1]),d_{L^p})\) is not complete and its completion is the metric space \((L^p([0,1]),d_{L^p})\text{.}\) The topology of these spaces, for \(p\in(0,1)\text{,}\) does not come neither from a norm nor from a family of seminorms. These are called F-spaces.

Finally, notice that \(C^\infty([0,1])=\displaystyle\cap_{k=0}^\infty C^k([0,1])\text{.}\) Similarly, the set \(\displaystyle\cap_{p=1}^\infty L^p([0,1])\) is denoted by \(L^\infty([0,1])\) and its norm is
\begin{equation*} \|f\|_{L^\infty} = \lim_{p\to\infty} \|f\|_{L^p} = \sup\text{ess}_{x\in[0,1]}|f(x)|, \end{equation*}
where the sup has to be attained within a set of positive measure. \(L^\infty([0,1])\) is a Banach space but it is not separable. Notice that the completion of \(C^\infty([0,1])\) with respect to the norm \(\|\cdot\|_{L^\infty}\) is \(C^0([0,1])\text{,}\) that is a closed subspace of \(L^\infty([0,1])\text{.}\)

Ultimately, we have the following decreasing sequence of complete metric spaces, all Banach except for the \(C^\infty\text{,}\) that is Frechet, and for the \(L^p\) at the left of \(L^1\text{,}\) that are F-spaces, and all completions of \(C^\infty\) with respect to some norm or metric, except for \(L^\infty\text{:}\)
\begin{equation*} \dots\supset L^{1/2}\supset\dots\supset L^1\supset\dots\supset L^2\supset\dots\supset L^\infty\supset C^0\supset C^1\supset \dots\supset C^\infty. \end{equation*}
Moreover, all inclusions above are continuous. For instance, this means that if \(f\) and \(g\) are "\(C^0\)-close" then they are also "\(L^p\)-close" for each \(p>0\) and so on. The fact that "\(C^k\)-close" implies "\(C^{k'}\)-close" for every \(k'<k\) is obvious, since
\begin{equation*} \|f\|_{C^{k}}\leq\|f\|_{C^{k}}+\|f^k\|_{C^0}=\|f\|_{C^{k+1}}. \end{equation*}
Moreover, since the measure of \([0,1]\) is finite (and equal to 1), we also have that
\begin{equation*} \|f\|_{L^p}^p=\int_{[0,1]}|f(s)|^p\;ds\leq\|f\|_{C^0}. \end{equation*}
Finally, by Holder's inequality and, again, the fact that the measure of \([0,1]\) is finite, we have that (see Subsection 3.6.2)
\begin{equation*} \|f\|_{L^q}\leq\|f\|_{L^p}\text{ for }p\leq q. \end{equation*}

Subsection 3.1.2 Weak derivatives

The \(W^{k,p}\) norm. Let \(k>0\) and \(p\geq1\) and set
\begin{equation*} \|f\|_{W^{k,p}} = \|f\|_{L^{p}} + \|f'\|_{L^{p}} + \dots + \|f^{(k)}\|_{L^{p}}. \end{equation*}
This is a norm on \(C^\infty([0,1])\text{.}\) Let \(f_n\) be Cauchy for \(\|\cdot\|_{W^{k,p}}\text{.}\) Then \(f_n, f'_n,\dots,f^{(k)}_n\) are Cauchy for \(\|\cdot\|_{L^{p}}\) and so there are functions \(g_0,g_1,\dots,g_k\in L^p([0,1])\) such that, in \(L^p\text{,}\)
\begin{equation*} f_n\to g_0, f'_n\to g_1,\dots,f^{(k)}_n\to g_k. \end{equation*}
If \(g_0\in C^k([0,1])\text{,}\) then \(g_i=g^{(i)}\) for every \(i=1,\dots,k\text{.}\) If \(g_0\) is not so regular, then some of the \(g_i\) (possibly all of them) will not be the derivative of \(g_{i-1}\text{.}\)

In a way similar to how we did for \(L^p\text{,}\) it can be proved that the completion of \(C^\infty([0,1])\) with respect to \(\|\cdot\|_{W^{k,p}}\text{,}\) denoted by \(W^{k,p}([0,1])\text{,}\) is the space of all \(L^p\) functions that are "weakly differentiable" \(k\) times and whose first \(k\) weak derivatives are \(L^p\text{.}\) The spaces \(W^{k,p}([0,1])\) for \(p=2\) are usually denoted by \(H^k([0,1])\text{.}\)
Consider the sequence of smooth functions
\begin{equation*} f_n(x)=\sqrt{(x-0.5)^2+\frac{1}{n}}\in C^\infty([-1,1]). \end{equation*}
We saw above (see around Eq. (3.1.1)) that \(f_n\) converges to \(g\) in \(C^0\text{.}\) Since \(\|f\|_{L^p}\leq\|f\|_{C^0}\text{,}\) this means that \(f_n\) converges to \(g\) in \(L^p\) for every \(p>0\text{.}\) Clearly \(g_0\) is not \(C^1\) but the sequence
\begin{equation*} f'_n(x)=\frac{x}{\sqrt{x^2+\frac{1}{n}}} \end{equation*}
converges in \(L^p\text{,}\) for any \(p>0\text{,}\) to the function \(g_1(x)\) that is \(-1\) at the left of 0.5 and \(+1\) at its right. Indeed \(f'_n\) converges pointwise to \(g_1\) at every \(x\neq0.5\) and \(|f'_n(x)|\leq1\) in \([0,1]\text{,}\) so that
\begin{equation*} |f'_n(x)-g_1(x)|^p\leq(|f'_n(x)|-|g_1(x)|)^p\leq(1+1)^p=2^p. \end{equation*}
Hence, by the dominated convergence theorem (see ???), \(\|f'_n-g_1\|_{L^p}\to0.\) This means that \(g_0\in W^{1,p}([0,1])\) and that \(g_1\) is its weak derivative.
The sequence of spaces \(W^{k,p}([0,1])\) is interwined with the sequence of the \(C^{k}([0,1])\text{.}\) For instance \(\cap_{k=1}^\infty W^{k,p}([0,1]) = C^\infty([0,1])\text{.}\) We will see this in more detail in the Chapter about Sobolev spaces.

Subsection 3.1.3 Schauder basis

The following is a natural extension to topological vector spaces of the concept of basis.
Definition 3.1.3. Schauder basis.
Let \(V\) be a Banach space. A Schauder basis is a sequence \(e_n\in V\) such that every element \(v\in V\) writes uniquely as a series
\begin{equation*} v = \sum_{n=0}^\infty v_n e_n \end{equation*}
for some sequence of scalars \(v_n\in\bK\text{.}\)
The sequence
\begin{equation*} \{1,\cos(x),\sin(x),\cos(2x),\sin(2x),\dots\} \end{equation*}
is a Schauder basis for all spaces \(L^p([0,2\pi])\text{,}\) \(p>1\) -- in other words, the Fourier series of a function converges to the function in the \(L^p\) norm for all \(p>1\text{.}\) On the contrary, it is known that in \(L^1([0,2\pi])\) there are functions whose Fourier series does not converge in the \(L^1\) norm.
Next example shows that the properties of being dense and to be a basis do not coincide.
The sequence
\begin{equation*} \{1,x,x^2,\dots\} \end{equation*}
is not a Schauder basis for \(C^0([0,1])\) because, although for every \(f\in C^0([0,1])\) and \(\eps>0\) one can find a polynomial \(p_\eps\) with \(\|f-p_\eps\|<\eps\text{,}\) \(p_\eps\) is not converging to a power series as \(\eps\to0\) (i.e. its coefficients do not change continuously with \(\eps\)).

Another way of seeing this is that functions that are power series are, by definition, analytic and clearly there are continuous functions that are not analytic. Notice, though, that, if a function is a power series, then the coefficients of its expansion in the monomials \(x^n\) are unique since the only power series equal to zero is the power series with all coefficients equal to zero.
Unlike the finite-dimensional case, though, there are separable Banach spaces that have no Scahuder basis (see ???).

Subsection 3.1.4 Some deep result on Banach spaces

Banach spaces can be quite complicated. For instance, not all separable Banach spaces have a Schauder basis. One might think that such complicated behavior are found only in abstract artificial cases. The first result shows that it is not so: The second result shows an important characterization of which Banach spaces are also Hilbert spaces (see next Section).
Definition 3.1.7. Complemented subspace.
A closed linear subspace \(M\) of a topological vector space \(V\) is complemented if there is a second closed linear subspace \(N\) of \(V\) such that \(M\oplus N=V\text{.}\)
The third result shows an important characterization of which Banach spaces are reflexive (see Section ???).
Definition 3.1.9. Proximinal subset.
A closed convex subset \(M\) of a Banach space \(B\) is proximinal if, for every \(x\in B\text{,}\) there is a \(y\in M\) such that \(d(x,M)=d(x,y)\text{.}\)
Closed linear subspaces of a Banach space are not necessarily proximal. We will see below that that this is instead the case for Hilbert spaces.
Let \(M\) be the subset of \(C^0([-1,1])\) of all functions such that
\begin{equation*} \int_{[-1,0]}f(x)dx = \int_{[0,1]}f(x)dx=0 \end{equation*}
One can verify easily that \(M\) is a closed, linear subspace of \(C^0([-1,1])\text{.}\)

Let \(g\) be the identity function \(g(x)=x\text{,}\) so that
\begin{equation*} -\int_{[-1,0]}g(x)dx = \int_{[0,1]}g(x)dx=1/2. \end{equation*}
Then, for each \(f\in M\text{,}\)
\begin{equation*} \int_{[-1,0]}(f(x)-g(x))dx - \int_{[0,1]}(f(x)-g(x))dx = 1 \end{equation*}
and so \(M\) has an element closest to \(g\) if and only if the affine subspace \(A\subset C^0([-1,1])\) of all functions \(h\) such that
\begin{equation*} \int_{[-1,0]}h(x)dx - \int_{[0,1]}h(x)dx = 1 \end{equation*}
has an element of minimal norm (i.e. minimal distance from the origin).

To see that this is not possible, first notice that \(d(A,0)\geq\|h\|_{C^0}\geq1/2\) for \(h\in A\) since
\begin{equation*} \left|\int_{[a,b]}h(x)dx\right|\leq|b-a|\cdot\|h\|_{C^0} \end{equation*}
and so
\begin{equation*} 1 = \int_{[-1,0]}h(x)dx - \int_{[0,1]}h(x)dx \leq 2\|h\|_{C^0}. \end{equation*}
Moreover, \(d(A,0)=1/2\text{.}\) Indeed, let \(f_n\) be
\begin{equation*} f_n(x)=\begin{cases}1\amp,1\in[-1,-1/n]\\ -nx\amp,x\in[-1/n,1/n]\\ -1\amp,x\in[1/n,1]\end{cases}, \end{equation*}
Then the sequence \(c_nf_n\text{,}\) where \(c_n=\frac{1}{2(1-1/n+1/n^2)}\text{,}\) lies in \(A\) and
\begin{equation*} d(A,0)\leq\lim_{n\to\infty}\|c_nf_n\|_{C^0}=1/2. \end{equation*}
There is no function \(h\in A\text{,}\) though, whose norm is \(1/2\) since then the integral from \(-1\) to 0 would be no more than \(1/2\) and the integral from \(0\) to \(1\) could be no less than \(-1/2\text{.}\) In order for both these conditions to hold, we should have that \(\lim_{x\to0^-}h(x)=1/2\) and \(\lim_{x\to0^+}h(x)=-1/2\text{,}\) against the hypothesis that \(h\in C^0([-1,1])\text{.}\)
Notice that there can be also closed linear subspaces where there are infintiely many points that achieve minimal distance, as next example shows.
Consider the subspace \(Z\subset L^1([0,1])\) of all Lebesgue-integrable functions such that \(\int_{[0,1]}f(x)dx=0\) and let \(g(x)=1\text{.}\) Then there is an element \(f\in Z\) with \(d(f,g)=d(Z,g)\) if and only if the affine subspace \(A\) of all \(h\in L^1([0,1])\) with \(\int_{[0,1]}h(x)dx=1\) has an element of minimal norm.

Since
\begin{equation*} 1 = \int_{[0,1]}h(x)dx\leq\int_{[0,1]}|h(x)|dx=\|h\|_{L^1}, \end{equation*}
then \(d(Z,0)\geq1\) and clearly \(d(Z,0)=1\) since \(d(1,0)=1\text{.}\) Moreover, every positive function whose integral is 1 belongs to \(Z\) and has distance 1 from the origin.